Problem 20 - University Physics with Modern Physics

Question

A 4.80-kg watermelon is dropped from rest from the roof of an 18.0-m-tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what is the watermelon's (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be \(different\) if there were appreciable air resistance?

Step-by-Step Solution

Answer

The work done by gravity is 847.44 J; speed before impact is 18.79 m/s. With air resistance, speed and kinetic energy would be lower.

Step 1: Understanding Work Done by Gravity

The work done by gravity is calculated using the formula \( W = mgh \), where \( m \) is the mass of the watermelon, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( h \) is the height from which the watermelon is dropped (18.0 m). Substitute the known values into the formula: \( W = 4.80 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 18.0 \, \text{m} \).

Step 2: Calculating Work Done by Gravity

Calculate the work done: \( W = 4.80 \times 9.81 \times 18.0 = 847.44 \, \text{Joules}.\) Therefore, the work done by gravity on the watermelon is \( 847.44 \, \text{J} \).

Step 3: Finding Kinetic Energy Before Impact

The kinetic energy (KE) of the watermelon just before striking the ground can be found using the principle of conservation of energy. Initially, all energy is potential energy: \( PE = mgh \). As the watermelon falls, this potential energy is converted into kinetic energy, so \( KE = mgh \). Calculate \( KE \) using \( KE = 847.44 \, \text{J} \).

Step 4: Finding the Speed Before Impact

The speed just before impact can be determined using the kinetic energy formula: \( KE = \frac{1}{2}mv^2 \). Rearrange to solve for \( v \): \( v = \sqrt{\frac{2 \cdot KE}{m}} \). Substitute the known values: \( v = \sqrt{\frac{2 \times 847.44}{4.80}} \).

Step 5: Calculating the Speed Before Impact

Calculate the speed: \( v = \sqrt{\frac{1694.88}{4.80}} = \sqrt{353.1} = 18.79 \, \text{m/s}.\) Thus, the speed of the watermelon just before impact is approximately \( 18.79 \, \text{m/s} \).

Step 6: Analyzing the Effect of Air Resistance

If there were appreciable air resistance, the work done by gravity would still be the same, but the kinetic energy and speed just before impact would be lower because some of the gravitational energy would be dissipated as heat due to air friction.

Key Concepts

Work Done by Gravity

Gravity is a conservative force that acts on objects, doing work whenever there is a vertical displacement. The work done by gravity on an object can be calculated using the formula \[ W = mgh \]where:

• \( m \) is the mass of the object (in kilograms).
• \( g \) represents the acceleration due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \) on the Earth's surface.
• \( h \) is the height or vertical distance through which the object falls (in meters).

For the watermelon dropped from a height of 18 meters, with a mass of 4.80 kg, the work done by gravity is calculated by substituting these values into the formula:\[ W = 4.80 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 18.0 \, \text{m} \]The resulting value is \( 847.44 \, \text{Joules} \), indicating that gravity performs this amount of work on the watermelon as it falls to the ground. This work translates into the energy that is transformed as the object descends.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. When an object falls under the influence of gravity from a height, the potential energy it initially holds is transformed into kinetic energy as it picks up speed. This can be expressed as:\[ KE = \frac{1}{2}mv^2 \]where:

• \( KE \) is the kinetic energy (in Joules).
• \( m \) denotes the mass of the object (in kilograms).
• \( v \) signifies the velocity of the object (in meters per second).

In our example of the watermelon, just before it hits the ground, all its potential energy has been converted into kinetic energy. Therefore, the kinetic energy is equal to the work done by gravity \( 847.44 \, \text{J} \). This allows us to solve for the speed before impact. Using the rearranged formula:\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]we find that:\[ v = \sqrt{\frac{2 \times 847.44}{4.80}} = 18.79 \, \text{m/s} \]Thus, the speed of the watermelon right before it touches the ground is approximately 18.79 meters per second.

Effect of Air Resistance

Air resistance, also known as drag, acts against the motion of falling objects, reducing their speed as they move through the air. If significant air resistance were present, it would affect the kinetic energy and speed of the watermelon just before impact.

• Without air resistance, all the potential energy converts into kinetic energy.
• With air resistance, some energy is lost as heat and sound as the watermelon falls.

Hence, the presence of appreciable air resistance would lower both the kinetic energy and the speed of the watermelon just before it hits the ground. The value of work done by gravity remains the same regardless of air resistance, but the redistribution of energy changes because some of it is used to overcome air resistance.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In the context of projectile motion, like our falling watermelon, energy transforms from potential to kinetic as it falls.

• Initially, the watermelon has maximum potential energy at the top of the building.
• As it descends, potential energy decreases as it's converted into kinetic energy.
• At the point just before impact, all potential energy is converted to kinetic energy (assuming no air resistance).

This transformation demonstrates the conservation of energy where the total energy in the system remains constant, provided there is no energy loss to non-conservative forces like air resistance. If air resistance is present, conservation of energy still holds, but accounting for lost energy in the form of heat is necessary, indicating that not all potential energy converts to kinetic energy.